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Question : Let $p(z)$ be a polynomial of degree n satisfying $|p(z)|\leq 1$ for all $z, |z| \leq 1$. Show that then $|p(z)| \leq |z|^n$ for all $z, |z| >1$.

This is what I could get done :- Let $p(z) = \sum_{k=0}^{n-1} c_k z^k + z^n $. Given $|p(z)| \leq 1$ when $|z|=1$. Then by Cauchy's estimates , $|c_k| = \frac{|p^{(k)}(0)|}{k!} \leq 1$ for all $k$. Therefore, $|p(z)| \leq \sum_{k=0}^{n-1} |c_k| |z|^k + |z|^n \leq \sum_{k=0}^{n-1} |z|^k + |z|^n \leq \dfrac{|z|^{n+1}-1}{|z|-1}$. From this we can say that when $|z| >> 1$, then $|p(z)| \leq |z|^n$. But how do I complete the argument for all $z \in \{|z|>1 \} $?

Sourav D
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  • I know that $|p(z)| = 1$ for at least one point when $|z| =1$. Also by maximum modulus principle, this leads to $|p(z)|<1$ when $z \in \mathbb{D}$. – Sourav D Dec 13 '13 at 01:39
  • Also for any circle $|z|>1$, $|p(z)|>1$ since $p(z)$ is entire, $|p(z)| = 1$ somewhere on the unit circle and maximum modulus principle applies to all domains $|z|\leq r$ where $r>1$. – Sourav D Dec 13 '13 at 01:50
  • I just found out that this question has already been answered, quite elegantly, here. http://math.stackexchange.com/questions/598135/complex-polynomial-of-degree-n?lq=1 – Sourav D Dec 13 '13 at 09:14

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